
 If \( f :=\frac{1}{2} \Vert \cdot \Vert ^{2}\) , in view of cor:characterCCS:rightcor:characterCCS:right:eq, we have \(\overleftarrow{CC{}}_{f}(\mathcal {S}) =\overrightarrow{CC{}}_{f}(\mathcal {S}) =\overleftarrow{CC{}}^{ps}_{f}(\mathcal {S}) =\overrightarrow{CC{}}^{ps}_{f}(\mathcal {S})=CC{(\mathcal {S})} = (1, \frac{3}{2}, \frac{3}{2})^{\intercal }\) .

 Suppose that \((\forall x \in \left[0, +\infty \right[^{3} )\)  \(f(x) =\sum ^{3}_{i=1} x_{i} \ln ( x_{i}) -x_{i}\) . Then \(f(1,1,1)=-3\) , \(f(1,2,1)=2\ln (2) -4\) , \(f(1,1,2)=2\ln (2) -4\) . Moreover, \((\forall x \in \left]0, +\infty \right[^{3} )\)  \(\nabla f(x) = \left( \ln (x_{1}), \ln (x_{2}), \ln (x_{3}) \right)^{\intercal } \) , and, in view of [1]} and [2]}, \((\forall x \in \mathbb {R}^{3} )\)  \(f^{*}(x) =\sum ^{3}_{i=1} \exp (x_{i} )\)  and \(\nabla f^{*}(x) =\left( \exp (x_{1} ), \exp (x_{2} ), \exp (x_{3} ) \right)^{\intercal }\) .

 exam:Back:Rnexam:Back:Rn:negative: and exam:forward:Rnexam:forward:Rn:negative: imply that
\(\overleftarrow{CC{}}_{f}(\mathcal {S}) = \left\lbrace  \left(1, \frac{1}{\ln 2} , \frac{1}{\ln 2} \right) \right\rbrace  \quad \text{and} \quad \overrightarrow{CC{}}_{f}(\mathcal {S}) = \left\lbrace  \left(1, \frac{4}{\rm e} , \frac{4}{\rm e} \right) \right\rbrace .\) 

thm:unique:LinIndpPformula and thm:LinIndpPformula:TpseudoCCS imply that
\(\overleftarrow{CC{}}^{ps}_{f}(\mathcal {S}) = \left(0, \frac{1}{\ln 2}, \frac{1}{\ln 2} \right) \quad \text{and} \quad \overrightarrow{CC{}}^{ps}_{f}(\mathcal {S}) =\left( {\rm e}, \frac{4}{\rm e}, \frac{4}{\rm e} \right).\) 



